2/11 is 0.181818 as a decimal so closest to 0.
2.5x - 3 + 2x = 1.75x - 1.25x + 13
4.5x - 3 = 0.50x + 13
4.0x = 16
x = 4
Answer: 4
2. We'll assume those xs are in the numerator
(3/7)x - 1/4 + (3/7)x = (9/7)x - (3/7)x + 3/4 - 1
(6/7)x - 1/4 = (6/7)x - 1/4
0 = 0
That's always true.
Answer: Any x is a solution.
(13•4) + (3•8) +3 = 79
The total cost is $79
Answer:
Each of the other cats have 45 whiskers on average.
Step-by-step explanation:
Let x represent the average number of whisker of each cat.
We have been given that there are 7 cats in my neighborhood, with an average of 41 whiskers each.
The total number of whiskers of six cats would be
.
Since one of the cats has 17 whiskers, so the total number of whiskers of 7 cats would be 
We will use average formula to solve our given problem.

Upon substituting our given values, we will get:

Let us solve for x.




Switch sides:



Therefore, the each of the other cats have 45 whiskers on average.
Answer:
about 78 years
Step-by-step explanation:
Population
y =ab^t where a is the initial population and b is 1+the percent of increase
t is in years
y = 2000000(1+.04)^t
y = 2000000(1.04)^t
Food
y = a+bt where a is the initial population and b is constant increase
t is in years
b = .5 million = 500000
y = 4000000 +500000t
We need to set these equal and solve for t to determine when food shortage will occur
2000000(1.04)^t= 4000000 +500000t
Using graphing technology, (see attached graph The y axis is in millions of years), where these two lines intersect is the year where food shortages start.
t≈78 years